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Talk:Bentley's number
What is this number meant to represent in the story? I assume it was meant to be the total count of wheels on all ten counters. If this is so, a careful reading of the story reveals that the value listed here can not be correct. Recall that the first counter contains 1 wheel, and each succeeding counter has a number of wheels that is a power of 10. It follows then that the count of all wheels must end in a 1, yet the number listed here must end in a zero. Remember the first counter has 1 wheel, and can be in 10 states, the second counter has 10 wheels and can be in 10,000,000,000 states, the third counter has 10,000,000,000 wheels and can be in 10^10^10 or'' trialogue'' states, the fourth counter has a'' trialogue'' wheels and can be in 10^^4 or a tetralogue states, etc. In general the nth counter has 10^^(n-1) wheels and 10^^n states. Therefore the tenth counter must have 10^^9 or an'' ennalogue wheels and 10^^10 or a ''dekalogue states. So Bentley's number should be redefined as (10^^0)+(10^^1)+(10^^2)+ ... +(10^^7)+(10^^8)+(10^^9), where b^^0 := 1. If the summation indexes were changed from i=1,10 to i=0,9, it would give the total number of wheels on all ten counters. If anyone agrees that this number is incorrectly defined perhaps it could be changed. 18:50, January 27, 2012 (UTC) anon. You're right. I'll correct it LittlePeng9 (talk) 16:38, February 23, 2013 (UTC) Should we use the original definition, or the corrected definition? -- I want more 10:36, April 25, 2013 (UTC) :The corrected one. I goofed up badly while making this. FB100Z • talk • 18:41, April 25, 2013 (UTC) Other meaning of “Bentley” “Bentley” is an automobile brand; automobiles are equipped with an odometer; and furthermore, the story Forever Endeavor, on which Bentley's Number is based, features odometer-like counters. -- 20:02, January 26, 2014 (UTC) Can wa add an image of a Bentley automobile to the article? -- 14:49, January 31, 2018 (UTC) Prime or composite? I am not sure about the primality of Bentley's number. -- 08:53, September 26, 2015 (UTC) : If my program is correct, then 347 is a factor of Bentley's number, so it must be composite. LittlePeng9 (talk) 20:38, September 26, 2015 (UTC) :: how'd you find that out? Cookiefonster (talk) 20:48, September 26, 2015 (UTC) :::I have written a C++ program which computed \(10\uparrow\uparrow i\mod n\) by repeated applications of Euler's theorem and modular exponentiation. Tomorrow I will go through the code once again to see if I didn't make a mistake (which I might have, because next divisor which the program found is 596219, which should be prime but isn't; that might be a problem due to rounding), and then I can post it for you to see. LittlePeng9 (talk) 20:58, September 26, 2015 (UTC) Bad news and good news: The bad news is that there was a mistake in my program, caused by using Euler's phi function when it could not be used, and also I think for some reason I was subtracting 1 when computing the residue modulo a number. Turns out 347 is not a factor of Bentley's number. Good news is that I have written a program which almost certainly is correct, and it has found 659 and 1873 to be prime factors of Bentley's number, and moreover they are the only prime factors below 1000000. For fun I'm running the computation up to \(10^7\) now. Here is my code written in Sage if anyone cares enough to check it. def ppow(k,n): #function computing 10^^k (mod n) if k 0: return 1 if k 1: return 10%n if k 2: return power_mod(10,10,n) if k 3: return power_mod(10,10^10,n) A = prime_to_m_part(n,10) #splitting n into part A, relatively prime to 10, and remaining part B B = n//A ex = ppow(k-1,euler_phi(A)) #when computing modulo A, we only have to care about exponent modulo phi(A) thanks to Euler's theorem X = power_mod(10,ex,A) return crt(X,0,A,B) #Chinese remainder theorem application; we may assume the tower is 0 mod B because the cases with exponent smaller than 10^10 are covered above def Bent(p): #Bentley's number modulo p return sum(for k in range(10))%p for p in prime_range(10000000): #a loop searching for prime factors if p%100000<100: 'beep',p//100000 #beep if Bent(p) 0: p (the line with "beep" is there only so that I know how far the computation is) LittlePeng9 (talk) 19:40, November 30, 2016 (UTC) :Update: no other prime divisors below \(10^7\). Also, no divisors below \(10^7\) apart from \(1,659,1837,659\cdot 1837\). Plus, added some comments to the code. LittlePeng9 (talk) 19:52, November 30, 2016 (UTC) Placement on the list Shouldn't this be above Decker? :No, this number is smaller than Decker. LittlePeng9 (talk) 13:04, January 21, 2017 (UTC)